Posted by hannah on Saturday, December 11, 2010 at 9:03am.
Some sulfuric acid is spilled on a lab bench. It can be neutralized by sprinkling sodium bicarbonate on it and then mopping up the resultant solution.
The sodium bicarbonate reacts with sulfuric acid as follows.
2 NaHCO3(s) + H2SO4(aq) --> Na2SO4(aq) + 2 H2O(l) + 2 CO2(g)
Sodium bicarbonate is added until the fizzing due to the formation of CO2(g) stops. If 34 mL of 6.0 M H2SO4 was spilled, what is the minimum mass of NaHCO3 that must be added to the spill to neutralize the acid?
i got 5.7 but it said:
"Your response differs from the correct answer by more than 10%. Double check your calculations."
what am i doing wrong?
chem - Dr Russ, Saturday, December 11, 2010 at 10:45am
The equation tells you that 2 moles of bicarbonate react with each mole of H2SO4
number of moles of H2SO4
6.0 mole L^-1 x 0.034 L= 0.204 mole
so 0.408 moles of bicarbonate are needed.
The molar mass of NaHCO3 is
84 g mole^-1
so mass of NaHCO3
0.408 mole x 84 g mole^-1 =34 g
2 sig figs appropriate in the answer as both starting values are to 2 sig figs.
chem - hannah, Saturday, December 11, 2010 at 2:15pm
i don't think 34 is correct because that is way more than 10% of my first answer of 5.7
chem - DrBob222, Saturday, December 11, 2010 at 2:34pm
I worked the same problem for another student a day or so ago and came up with approximately 34g as I remember. And if you will look at the response to your 5.7 value, it says "Your response differs from the correct answer by more than 10%. Double check your calculations." The bold faced print is mine. So why can't 34 be correct? It differs, not by 10% which you imply it should, but by MORE THAN 10%, which it does.
chem - Anonymous, Tuesday, February 17, 2015 at 3:41pm
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